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Difference between revisions of "2016 AMC 12A Problems/Problem 9"

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==Solution 2==
 
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The diagonal of the small square can be written in two ways: <math>s \sqrt(2)</math> and <math>2*(1-2s).</math> Equating and simplifying gives <math>4 + 7 \Rightarrow \boxed{\textbf{(E) } 11}.</math>
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The diagonal of the small square can be written in two ways: <math>s \sqrt(2)</math> and <math>2*(1-2s).</math> Equating and simplifying gives <math>s = \frac{4 - \sqrt{2}}{7}</math>. Hence our answer is <math>4 + 7 \Rightarrow \boxed{\textbf{(E) } 11}.</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=A|num-b=8|num-a=10}}
 
{{AMC12 box|year=2016|ab=A|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:02, 22 January 2019

Problem 9

The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is $\tfrac{a-\sqrt{2}}{b}$, where $a$ and $b$ are positive integers. What is $a+b$ ?

[asy] real x=.369; draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); filldraw((0,0)--(0,x)--(x,x)--(x,0)--cycle, gray); filldraw((0,1)--(0,1-x)--(x,1-x)--(x,1)--cycle, gray); filldraw((1,1)--(1,1-x)--(1-x,1-x)--(1-x,1)--cycle, gray); filldraw((1,0)--(1,x)--(1-x,x)--(1-x,0)--cycle, gray); filldraw((.5,.5-x*sqrt(2)/2)--(.5+x*sqrt(2)/2,.5)--(.5,.5+x*sqrt(2)/2)--(.5-x*sqrt(2)/2,.5)--cycle, gray); [/asy]

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution

Let $s$ be the side length of the small squares.

The diagonal of the big square can be written in two ways: $\sqrt{2}$ and $s \sqrt{2} + s + s \sqrt{2}$.

Solving for $s$, we get $s = \frac{4 - \sqrt{2}}{7}$, so our answer is $4 + 7 \Rightarrow \boxed{\textbf{(E) } 11}$

Solution 2

The diagonal of the small square can be written in two ways: $s \sqrt(2)$ and $2*(1-2s).$ Equating and simplifying gives $s = \frac{4 - \sqrt{2}}{7}$. Hence our answer is $4 + 7 \Rightarrow \boxed{\textbf{(E) } 11}.$

See Also

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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